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a(n) = 111111 in base n.
14

%I #38 Oct 04 2024 11:18:51

%S 6,63,364,1365,3906,9331,19608,37449,66430,111111,177156,271453,

%T 402234,579195,813616,1118481,1508598,2000719,2613660,3368421,4288306,

%U 5399043,6728904,8308825,10172526,12356631,14900788,17847789,21243690,25137931

%N a(n) = 111111 in base n.

%H Carlos M. da Fonseca and Anthony G. Shannon, <a href="https://doi.org/10.7546/nntdm.2024.30.3.491-498">A formal operator involving Fermatian numbers</a>, Notes Num. Theor. Disc. Math. (2024) Vol. 30, No. 3, 491-498.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).

%F a(n) = n^5 + n^4 + n^3 + n^2 + n + 1 = (n^6-1)/(n-1).

%F G.f.: x*(6 + 27*x + 76*x^2 + 6*x^3 + 6*x^4 - x^5)/(1-x)^6. - _Colin Barker_, May 08 2012

%F E.g.f.: exp(x)*(1 + 5*x + 26*x^2 + 32*x^3 + 11*x^4+ x^5) - 1. - _Stefano Spezia_, Oct 03 2024

%e a(3)=364 because 111111 base 3 = 243 + 81 + 27 + 9 + 3 + 1 = 121.

%t Table[(n^5+n^4+n^3+n^2+n+1),{n,0,5!}] (* _Vladimir Joseph Stephan Orlovsky_, Mar 06 2010 *)

%o (PARI) a(n)=(n^6-1)/(n-1) \\ _Charles R Greathouse IV_, Sep 28 2015

%Y 6th row of the array A055129.

%Y Cf. A104878.

%K base,nonn,easy

%O 1,1

%A _Henry Bottomley_, Mar 23 2000